SOLUTION: Here's an equation:
x=sqrt(15-2x)
This is how I solved it:
x^2=15-2x
x^2+2x-15=0
(x+5)*(x-3)=0
therefore x=-5 or x=3 and after checking I found that both solutions sati
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Radicals
-> SOLUTION: Here's an equation:
x=sqrt(15-2x)
This is how I solved it:
x^2=15-2x
x^2+2x-15=0
(x+5)*(x-3)=0
therefore x=-5 or x=3 and after checking I found that both solutions sati
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Question 596100: Here's an equation:
x=sqrt(15-2x)
This is how I solved it:
x^2=15-2x
x^2+2x-15=0
(x+5)*(x-3)=0
therefore x=-5 or x=3 and after checking I found that both solutions satisfies the original equation [because 3=sqrt(9) and -5=sqrt(25)]
But after checking the answers, I found that -5 is considered as an extraneous root as -5 is not equal to sqrt(25), namely 5. But the square root of 25 can be both 5 or -5.
Can anyone tell me why? Answer by jim_thompson5910(35256) (Show Source):
You can put this solution on YOUR website! Actually, the square root of 25 is only 5 because the square root function is what it says it is: a function.
A function takes one input and spits out only ONE output. So it can't take in the input 25 and spit out both 5 and -5 at the same time.
So
When we want the plus/minus to be in there, we have to explicitly add it like so
(